Your goal is to reach the last index in the minimum number of jumps.

For example:
Given array A = [2,3,1,1,4]

The minimum number of jumps to reach the last index is 2. (Jump 1 step from index 0 to 1, then 3 steps to the last index.)

選擇最長的步數能保證最小的次數。

class Solution {

public:

    int jump(int A[], int n) {

        if(0==n)

           return 0;

        int cur=0,next=0,maxdistance=-1,count=0;//初始化老是出錯

        while(true)

        {

            if(next>=n-1) //寫成maxdistance>=n-1老是[0]測試用例通不過

               return count;

            for(cur;cur<=next;++cur)

            {

                maxdistance=max(maxdistance,cur+A[cur]);//貪心演算法,選擇cur到next範圍內最大的移動值，當前編號+當前位置可移動的最大值

            }

cur=next; //下一步的新起點

            next=maxdistance; //這一步能夠走得最大值

            count++;

        }

    }

};

題目2：

Say you have an array for which the i^th element is the price of a given stock on day i.

If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.

找到最大利潤，你只能進行一次交易

分析：

貪心法，分別找到價格最低和最高的一天，低進高出，注意最低的一天要在 把原始價格序列變成差分序列，本題也可以做是最大 m 子段和，m = 1。

i--版本 用最大值

class Solution {

public:

    int maxProfit(vector<int> &prices) {

        if(prices.size() == 0)

            return 0;

        int ans;//最大差價

        int maxPrices=prices[prices.size()-1];

        for(int i=prices.size()-1;i>=0;i--)

        {

            maxPrices=max(maxPrices,prices[i]);//第i天之後,即i+1到第n天的最大利益maxPrices,即找曲線的最大值

            ans=max(ans,maxPrices-prices[i]);//即i+1到第n天的最大利益maxPrices減去第i天，即第i天最大利益;ans為前面i天中利益最大的一天

        }

        return ans;

    }

};

i++ 版本 用最小值

Say you have an array for which the i^th element is the price of a given stock on day i.

Design an algorithm to find the maximum profit. You may complete as many transactions as you like (ie, buy one and sell one share of the stock multiple times). However, you may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

找到最大利潤，你可以進行多次交易 不需要最大最小值，只要交易能夠賺錢，我就交易。 class Solution { public:     int maxProfit(vector<int> &prices) {         if(prices.size()==0 || prices.size()==1)             return 0;         int ans[prices.size()-1];         int buypoint=0,sellpoint=prices.size()-1;         ans[0]=0;         for(int i=0;i<prices.size()-1;i++)         {             if(prices[i+1]>=prices[i])             {                 sellpoint=i+1;                 buypoint=i;//第一次寫忘了             }             else             {                 ans[i+1]=ans[i]; /*如果prices[i+1]<prices[i]，前i+1天最大利益儲存不變，同時選擇最低點買入股票*/                 buypoint=i+1;             }             if(sellpoint>=buypoint && prices[i+1]>=prices[i])//第一次寫這裡也忘了              ans[i+1]=max(prices[sellpoint]-prices[buypoint],ans[i]+prices[i+1]-prices[i]);         }         return ans[prices.size()-1];     } }; 簡化版本： 分析 貪心法，低進高出，把所有正的價格差價相加起來。 把原始價格序列變成差分序列，本題也可以做是最大 m 子段和，m = 陣列長度。 程式碼 // LeetCode, Best Time to Buy and Sell Stock II // 時間複雜度 O(n)，空間複雜度 O(1) class Solution { public: int maxProfit(vector<int> &prices)  {     int sum = 0;     for (int i = 1; i < prices.size(); i++)      {         int diff = prices[i] - prices[i - 1];         if (diff > 0) sum += diff;     }     return sum;     } }; 題目3： Longest Substring Without Repeating Characters 描述 Given a string, find the length of the longest substring without repeating characters. For example, the longest substring without repeating letters for ”abcabcbb” is ”abc”, which the length is 3. For ”bbbbb” the longest substring is ”b”, with the length of 1.

題4：

Container With Most Water

描述

Given n non-negative integers a1; a2; :::; an, where each represents a point at coordinate (i; ai). n verti-

cal lines are drawn such that the two endpoints of line i is at (i; ai) and (i; 0). Find two lines, which together

with x-axis forms a container, such that the container contains the most water.

Note: You may not slant the container